Chapter 9.8, Problem 2WE

To determine

To calculate: To find the real solutions

Answer to Problem 2WE

Solutions are $\left(40,7\right),\left(0,-3\right)$

Explanation of Solution

Given information: System of equations $x=y^2-9;x-4y-12=0$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}x=y^2-9-----(1)\\ x-4y-12=0---(2)\end{array}$

From equation (1),

  $x=y^2-9$

Substituting the value in equation (2)

  $\begin{array}{l}{y}^2-9-4y-12=0\\ y^2-4y-21=0\\ \left(y-7\right)\left(y+3\right)=0\\ y=7,-3\end{array}$

When $y=7$

  $\begin{array}{l}x={\left(7\right)}^2-9\\ x=49-9\\ x=40\end{array}$

When $y=-3$

  $\begin{array}{l}x={\left(-3\right)}^2-9\\ x=9-9\\ x=0\end{array}$

Conclusion:

Hence, solutions are $\left(40,7\right),\left(0,-3\right)$